Cardinality of a subset of [0,1]

[ribbon]

Homework Statement

What is the cardinality of the set of all numbers in the interval [0, 1] which
have decimal expansions with a finite number of non-zero digits?

Homework Equations

The Attempt at a Solution

I say its still c? Am I correct, there is no way I can pair this set with the natural numbers.

Define
f(1) = 0.1
f(2) = 0.02
f(3) = 0.003
f(4) = 0.0004
...
But then we must pair elements of N to 0.2, 0.3, 0.4... and 0.03, 0.04, 0.05, and 0.004, 0.005,...

This seems weak though no?

 

[micromass]

Your set is a subset of $\mathbb{Q}$, isn't it?

 

[verty]

@Ribbon: Your argument doesn't make any sense to me. You seem to be arguing that there are a countable number of sets equinumerous with the naturals. Is this what you are doing and what are you trying to achieve by it?

 

[tiny-tim]

hi ribbon!

…, there is no way I can pair this set with the natural numbers.

hint: can you pair the set of all numbers with two non-zero digits with the natural numbers?

 

[vela]

I say its still c? Am I correct, there is no way I can pair this set with the natural numbers.

Define
f(1) = 0.1
f(2) = 0.02
f(3) = 0.003
f(4) = 0.0004
...
But then we must pair elements of N to 0.2, 0.3, 0.4... and 0.03, 0.04, 0.05, and 0.004, 0.005,...

This seems weak though no?

Yes, your argument doesn't work. All you've shown is that you failed at your first attempt at finding a bijection between N and the set. Perhaps if you were more clever, you could find such a bijection.

 

[ribbon]

hi ribbon!


hint: can you pair the set of all numbers with two non-zero digits with the natural numbers?

Could I go,
f(1) = 0.11
f(2) = 0.12
f(3) = 0.13
...

and skip numbers like 0.2, 0.3? But then what happens to all the numbers in between I'm missing?

 

[ribbon]

Your set is a subset of $\mathbb{Q}$, isn't it?

It seems so, but the unit interval has cardinality c and at first glance it appeared to me that we were excluding a finite set from an infinite one, and if I am not mistaken with cardinal arithmetic, c would remain the cardinality. But it seems to be otherwise now.

 

[LCKurtz]

hi ribbon!


hint: can you pair the set of all numbers with two non-zero digits with the natural numbers?

Hey Tiny, what happened to your fish avatar? Did he try to swim through a screen?

 

[tiny-tim]

Could I go,
f(1) = 0.11
f(2) = 0.12
f(3) = 0.13
...

and skip numbers like 0.2, 0.3? But then what happens to all the numbers in between I'm missing?

but you're not even trying to count any numbers with more than 2 decimal places!

try again, but with binary numbers (instead of decimals), it's easier

Hey Tiny, what happened to your fish avatar? Did he try to swim through a screen?

it wasn't a proper underwater camera!

 

[ribbon]

but you're not even trying to count any numbers with more than 2 decimal places!

try again, but with binary numbers (instead of decimals), it's easier


it wasn't a proper underwater camera!

Hmm, ok well could the sequence be 0.1, 0.101, 0.010101... till I hit that n (n being the finite number of non zero digits?

 

[LCKurtz]

It seems so, but the unit interval has cardinality c and at first glance it appeared to me that we were excluding a finite set from an infinite one, and if I am not mistaken with cardinal arithmetic, c would remain the cardinality.

So, have you just walked away from the thread where we were discussing that topic?

 

[pasmith]

It seems so, but the unit interval has cardinality c and at first glance it appeared to me that we were excluding a finite set from an infinite one

Your set is infinite: if $x$ is in that set, then so is $10^{-n}x$ for all $n \in \mathbb{N}$. Adding any number of zeroes to the beginning of the decimal expansion does not change the number of non-zero digits in the expansion.

 

[HallsofIvy]

Homework Statement

What is the cardinality of the set of all numbers in the interval [0, 1] which
have decimal expansions with a finite number of non-zero digits?

You do not recognize this as the subset of all rational numbers whose denominators have only powers of 2 and powers of 5 as factors? If $$x= 0.a_1a_2\cdot\cdot\cdot a_n$$ has only "n" non-zero digits, then $$10^nx= a_1a_2\cdot\cdot\cdot a_n$$ and, finally, $$x= \frac{a_1a_2\cdot\cdot\cdot a_n}{10^n}$$. You may be able to do a lot of "cancelling" to reduce the denominator but 10= 2(5) so there will never be any factors other than "5" or "2" in the denominator.

Homework Equations

The Attempt at a Solution

I say its still c? Am I correct, there is no way I can pair this set with the natural numbers.

Define
f(1) = 0.1
f(2) = 0.02
f(3) = 0.003
f(4) = 0.0004
...
But then we must pair elements of N to 0.2, 0.3, 0.4... and 0.03, 0.04, 0.05, and 0.004, 0.005,...

This seems weak though no?